Question

A water cooler of storage capacity $120\text{litres}$ can cool water at a constant rate of $P$. In a closed circulation system (as shown schematically in this figure), the water from the cooler is used to cool an external device that constantly generates $3\text{kW}$ of power (thermal load). The temperature of the water fed into the device cannot exceed ${30}^{\circ }\text{C}$ and the entire $120\text{litres}$ of water is initially cooled to ${10}^{\circ }\text{C}$. The entire system is thermally insulated. The minimum value of $P$ for which the device can be operated for 3 hours (taking the specific heat capacity of water $=4.2{\text{kJ kg}}^{-1}{\text{K}}^{-1}$ and the density of water $=1000{\text{kg m}}^{-3})$ is

• A. $1600\text{W}$
• B. $2067\text{W}$
• C. $2533\text{W}$
• D. $3933\text{W}$

Answer 1

The correct option is B $2067\text{W}$

Given that,
Initial temperature, ${\theta }_1={10}^{\circ }\text{C}$
Final temperature, ${\theta }_2={30}^{\circ }\text{C}$
Mass of water in cooler $=120\text{kg}$
Change in temperature, $\Delta \theta ={20}^{\circ }\text{C}$

Heat generated by the device $=\left(3\text{kW)}\right(3\text{hrs})=3\times {10}^3\times 3\times 3600\text{J}$
$=324\times {10}^5\text{J}$
Heat absorbed by water $=m{c}_w\Delta \theta =\left(120\right)\left(4200\right)\left(20\right)=100.8\times {10}^5\text{J}$
Heat absorbed by the coolant,
$=Pt=(324\times {10}^5-100.8\times {10}^5)\text{J}$
$\Rightarrow P=\frac{223.2\times {10}^5\text{J}}{3\times 3600s}=2067\text{W}$
is the minimum rate of cooling.